题目内容
(1)
<
<
(2)
<
<
.
| 7 |
| 5 |
| 17 |
| () |
| 10 |
| 7 |
(2)
| 3 |
| 5 |
| () |
| 40 |
| 7 |
| 10 |
分析:(1)分子通分,可得
<
<
,依此可得( )为12;
(2)分母通分,可得
<
<
,依此可得( )的取值范围,从而求解.
| 1190 |
| 850 |
| 1190 |
| 70×() |
| 1190 |
| 833 |
(2)分母通分,可得
| 24 |
| 40 |
| () |
| 40 |
| 28 |
| 40 |
解答:解:(1)
<
<
则
<
<
,
则( )为12;
(2)
<
<
,
则
<
<
,
24<( )<28,则( )为27,26,25中任选一个.
| 7 |
| 5 |
| 17 |
| () |
| 10 |
| 7 |
则
| 1190 |
| 850 |
| 1190 |
| 70×() |
| 1190 |
| 833 |
则( )为12;
(2)
| 3 |
| 5 |
| () |
| 40 |
| 7 |
| 10 |
则
| 24 |
| 40 |
| () |
| 40 |
| 28 |
| 40 |
24<( )<28,则( )为27,26,25中任选一个.
点评:考查了分数的大小比较,本题(1)通过分子通分,(2)通过分母通分求解.
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